# AP Calculus BC Review Question 1

*A review by Omninox of an AP Calculus BC question*

Now we shift our focus from the most recent reviews of social subjects to mathematics. Specifically, we will cover a topic of Calculus BC. Now, BC has a mix of Calculus AB content and new BC material, but the majority of the questions deal with AB content. As a refresher, let's look at a problem from AB below:

**The graph below shows the velocity in cm/sec of a particle moving along the y -axis. The numbers in the graph are the areas of the enclosed regions. Note: Figure not drawn to scale!**

**The particle is at y = 1 at time t = a . What is the particle's distance from the origin at time t = e ?**

Your answer choices are:

- y = 4
- y = 0
- y = 1
- y = 5
- y = 3

The answer to this problem is **y=0**. In order to solve this question, you need to know two things: 1.) That velocity is the derivative result of distance and 2.) The integral of a graph is simply the area under a graph.

The question is asking for you to solve for the distance the particle travels from **point a** to **point e**, and you are given a velocity vs. time graph. Since velocity is the derivative of distance, you can solve for distance by taking the anti-derivative of velocity. The anti-derivative is the same as determining the integral of the graph. And as we mentioned earlier, the integral is simply the area under the curve. Well lucky for you, those numbers are already given to you in the graph above.

At **point a**, you start at a distance of y=1. From **a** to **c**, you gain a distance of 1, because that is the area under that section of the curve. From **c** to **d**, notice the curve goes under the x-axis. This indicates a negative value, or in other words, you're going backwards and losing distance. At **point d**, you lose a distance of 3, meaning the particle has travelled a total distance of -1 (y=2-3). From **d** to **e**, the area under the curve is positive 1, since the curve lies above the x-axis. This results in a net distance travelled of 0 (y =(-1) + 1). Therefore, the total distance from the origin of the particel from **point a to point e** is y=0.

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